Monthly Archives: June 2016

Consider two partially ordered sets \((E,\le)\) and \((F,\le)\) and a strictly increasing map \(f : E \to F\). If the order \((E,\le)\) is total, then \(f\) is one-to-one. Indeed for distinct elements \(x,y \in E\), we have either \(x < y\) or \(y < x\) and consequently \(f(x) < f(y)\) or \(f(y) < f(x)\). Therefore \(f(x)\) and \(f(y)\) are different.
This is not true anymore for a partial order \((E,\le)\). We give a counterexample.

Consider a finite set \(E\) having at least two elements and partially ordered by the inclusion. Let \(f\) be the map defined on the powerset \(\wp(E)\) that maps \(A \subseteq E\) to its cardinal \(\vert A \vert \). \(f\) is obviously strictly increasing. However \(f\) is not one-to-one as for distincts elements \(a,b \in E\) we have \[
f(\{a\}) = 1 = f(\{b\})\]

Consider a functions series \(\displaystyle \sum f_n\) of functions defined on a set \(S\) to \(\mathbb R\) or \(\mathbb C\). It is known that if \(\displaystyle \sum f_n\) is normally convergent, then \(\displaystyle \sum f_n\) is uniformly convergent.

A case where \(l=1\) and the series converges

Consider the series \(\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}\). We have \[\sqrt[n]{\frac{1}{n^2}} = \frac{1}{n^{\frac{2}{n}}}=e^{- \frac{2}{n} \ln n} \] Therefore \(\limsup\limits_{n \to \infty} \sqrt[n]{\frac{1}{n^2}} = 1\), while the series \(\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}\) is convergent as we have seen in the ratio test article. Continue reading Root test→

The ratio test is a test for the convergence of a series \[
\sum_{n=1}^\infty a_n \] where each term is a real or complex number and is nonzero when \(n\) is large. The test is sometimes known as d’Alembert’s ratio test.

Cases where \(l=1\) and the series diverges

Consider the harmonic series \(\displaystyle \sum_{n=1}^\infty \frac{1}{n}\). We have \(\lim\limits_{n \to \infty} \frac{n+1}{n} = 1\). It is well know that the harmonic series diverges. Recall that one proof uses the Cauchy’s convergence test based for \(k \ge 1\) on the inequalities: \[
\sum_{n=2^k+1}^{2^{k+1}} \frac{1}{n} \ge \sum_{n=2^k+1}^{2^{k+1}} \frac{1}{2^{k+1}} = \frac{2^{k+1}-2^k}{2^{k+1}} \ge \frac{1}{2}\]

An even simpler case is the series \(\displaystyle \sum_{n=1}^\infty 1\).

Cases where \(l=1\) and the series converges

We also have \(\lim\limits_{n \to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert = 1\) for the infinite series \(\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}\). The series is however convergent as for \(n \ge 1\) we have:\[
0 \le \frac{1}{(n+1)^2} \le \frac{1}{n(n+1)} = \frac{1}{n} – \frac{1}{n+1}\] and the series \(\displaystyle \sum_{n=1}^\infty \left(\frac{1}{n} – \frac{1}{n+1} \right)\) obviously converges.

Another example is the alternating series \(\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n}\).